You would need to look at all 20 sides to verify that one of the numbers hadn't been duplicated.posted by DWRoelands at 11:48 AM on November 29, 2012 [7 favorites]

You would have to inspect every face of the die to make sure that all the numbers from 1 to 20 are represented.posted by fantabulous timewaster at 11:54 AM on November 29, 2012

well obviously that :) I guess I was assuming that you knew there was one each. Logically you would have to look until there were only two numbers left, but I was wondering if someone had a "well statistically if they got x% correctly matched then the rest will be fine"posted by zombieApoc at 11:55 AM on November 29, 2012

If you can pick which numbers you look at, then you can look at just 18 (just make sure the remaining two add up to 21 so you don't care which is in which place). If you're just randomly being presented with faces, you'll probably have to look at 19. You can never get away with just looking at 17.posted by dfan at 12:08 PM on November 29, 2012 [1 favorite]

If you know there is one of each number from 1-20 on each face and are only checking to see that opposite faces sum to 21, then you need to check 18 of the faces (or 9 opposite-side pairs) to be sure that they all sum to 21.posted by Aizkolari at 12:08 PM on November 29, 2012 [1 favorite]

Given an n-sided die of standard shape, and assuming that each side has a whole number between 1 and n printed on it with each number being printed exactly once...

If we were to start examining the D-n, we would first look at one face and note what number is printed, there is a chance of (1/(n-1)) that the number on the opposite side is the correct one for a legal die.

If we noted these 2 faces did indeed add up to (n+1),and then viewed the third face, there is a (1/(n-3)) chance that the fourth face viewed is correct.

And so on, until (if all the previous faces were correct) we look at the (n-3)th face on the die, the chances of it's opposite being correct are (1/n-(n-3)) or (1/3).

So, probability-wise, if you got (n-2/n) faces correct, there's a 1/1 chance rest would be fine, but that's only a different way of saying you'd have to look until there are two numbers left.

The making of a probability table is left to the reader.

Statistically, OTOH... there are some really nice steampunk dice sets on Amazon. If you'd like to buy me 10 sets or so of them in various colors, I will happily look at them and make some statistical tables of how the faces are labeled.posted by yohko at 12:40 PM on November 29, 2012 [2 favorites]

So, probability-wise, if you got (n-2/n) faces correct, there's a 1/1 chance rest would be fine, but that's only a different way of saying you'd have to look until there are two numbers left.

This is only true if, as dfan points out, the two unseen faces are opposite faces. If the two unseen faces are not opposite each other, having seen the other n-2 is not a guarantee that the final two are in the correct places.

Is there an algorithm to state which numbers belong next to which other numbers?

No, because there are many possibilities for configurations which meet the stated requirements. For example, on a d20, any three numbers may appear adjacent next to the 1, except for 20 and as long as no two of those three add up to 21. There are 672 possibilities for which 3 numbers appear adjacent to the 1, and even that does not fully determine the configuration of the rest of the die.posted by DevilsAdvocate at 3:20 PM on November 29, 2012

as a side point, not all gaming dice follow the "opposite sides add to __" rule, either, so bear that in mind.posted by radiosilents at 5:31 PM on November 29, 2012

Statistical analysis of the likelihood of the unexamined sides being correct would only make sense if the numbers on the sides were being randomly generated. In the real world, this is not the case.

Either the D20 follows the standard numbering scheme used by the overwhelming majority of manufacturers or else it is a special die manufactured for an obscure game or customized the the user. If it is a special die, then you can't make any assumptions about the numbering unless you have some knowledge of the reason for its creation. For example, maybe a cheating D&D player might modify a D20 to have 2 20s and no 1s. In that case, you might not catch the bogosity until you examined all 20 sides.

Assuming that you have no information indicating a given D20 is likely to be non-standard, then the only statistical prediction you can make is based on the proportion of standard D20s to non-standard in the real world. In that case, without examining any sides at all you can make a prediction that the die will be a standard legal D20 with very close to 100% certainty.posted by tdismukes at 6:48 PM on November 29, 2012

This is only true if, as dfan points out, the two unseen faces are opposite faces. If the two unseen faces are not opposite each other, having seen the other n-2 is not a guarantee that the final two are in the correct places.

DevilsAdvocate, I refered to looking at opposite faces several times, and specified a die of standard shape (although I guess that doesn't work for a d-4, not having any opposite faces). I see now that I overlooked specifying that face 4 should be opposite face 3 -- you are correct that this cannot be assumed without that explicit specification. Thank you for pointing it out.

However, If you have viewed n-2 faces in groups of opposite pairs, the only 2 remaining faces will NOT have been opposite from any of the previously viewed faces, thus the only faces for them to be opposite from will be each other.

Standard shape is meant to exclude things like d-7 that have non-opposite faces, dice with curved faces, round dice, dice such as the ones in the linked comic, and the like.posted by yohko at 11:40 AM on November 30, 2012

OK, mea culpa on not noticing that you had indeed specified looking at opposite faces in pairs.

However, zombieApoc's question was not "if you randomly label the faces of a die [assuming a standard die, and also not a d4, which does have opposite faces] with the numbers 1 to n, not repeating any numbers, what is the probability that all pairs of opposite faces add to n+1?" That's what you seem to be answering, but it's not what was asked.posted by DevilsAdvocate at 3:51 AM on December 1, 2012

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